LeetCode #2959 — HARD

Number of Possible Sets of Closing Branches

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

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The Problem

Problem Statement

There is a company with n branches across the country, some of which are connected by roads. Initially, all branches are reachable from each other by traveling some roads.

The company has realized that they are spending an excessive amount of time traveling between their branches. As a result, they have decided to close down some of these branches (possibly none). However, they want to ensure that the remaining branches have a distance of at most maxDistance from each other.

The distance between two branches is the minimum total traveled length needed to reach one branch from another.

You are given integers n, maxDistance, and a 0-indexed 2D array roads, where roads[i] = [ui, vi, wi] represents the undirected road between branches ui and vi with length wi.

Return the number of possible sets of closing branches, so that any branch has a distance of at most maxDistance from any other.

Note that, after closing a branch, the company will no longer have access to any roads connected to it.

Note that, multiple roads are allowed.

Example 1:

Input: n = 3, maxDistance = 5, roads = [[0,1,2],[1,2,10],[0,2,10]]
Output: 5
Explanation: The possible sets of closing branches are:
- The set [2], after closing, active branches are [0,1] and they are reachable to each other within distance 2.
- The set [0,1], after closing, the active branch is [2].
- The set [1,2], after closing, the active branch is [0].
- The set [0,2], after closing, the active branch is [1].
- The set [0,1,2], after closing, there are no active branches.
It can be proven, that there are only 5 possible sets of closing branches.

Example 2:

Input: n = 3, maxDistance = 5, roads = [[0,1,20],[0,1,10],[1,2,2],[0,2,2]]
Output: 7
Explanation: The possible sets of closing branches are:
- The set [], after closing, active branches are [0,1,2] and they are reachable to each other within distance 4.
- The set [0], after closing, active branches are [1,2] and they are reachable to each other within distance 2.
- The set [1], after closing, active branches are [0,2] and they are reachable to each other within distance 2.
- The set [0,1], after closing, the active branch is [2].
- The set [1,2], after closing, the active branch is [0].
- The set [0,2], after closing, the active branch is [1].
- The set [0,1,2], after closing, there are no active branches.
It can be proven, that there are only 7 possible sets of closing branches.

Example 3:

Input: n = 1, maxDistance = 10, roads = []
Output: 2
Explanation: The possible sets of closing branches are:
- The set [], after closing, the active branch is [0].
- The set [0], after closing, there are no active branches.
It can be proven, that there are only 2 possible sets of closing branches.

Constraints:

  • 1 <= n <= 10
  • 1 <= maxDistance <= 105
  • 0 <= roads.length <= 1000
  • roads[i].length == 3
  • 0 <= ui, vi <= n - 1
  • ui != vi
  • 1 <= wi <= 1000
  • All branches are reachable from each other by traveling some roads.
Patterns Used
⚡Bit Manipulation→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: There is a company with n branches across the country, some of which are connected by roads. Initially, all branches are reachable from each other by traveling some roads. The company has realized that they are spending an excessive amount of time traveling between their branches. As a result, they have decided to close down some of these branches (possibly none). However, they want to ensure that the remaining branches have a distance of at most maxDistance from each other. The distance between two branches is the minimum total traveled length needed to reach one branch from another. You are given integers n, maxDistance, and a 0-indexed 2D array roads, where roads[i] = [ui, vi, wi] represents the undirected road between branches ui and vi with length wi. Return the number of possible sets of closing branches, so that any branch has a distance of at most maxDistance from any other.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Bit Manipulation

Example 1

3
5
[[0,1,2],[1,2,10],[0,2,10]]

Example 2

3
5
[[0,1,20],[0,1,10],[1,2,2],[0,2,2]]

Example 3

1
10
[]
Step 02

Core Insight

What unlocks the optimal approach

  • Try all the possibilities of closing branches.
  • On the vertices that are not closed, use Floyd-Warshall algorithm to find the shortest paths.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2959: Number of Possible Sets of Closing Branches
class Solution {
    public int numberOfSets(int n, int maxDistance, int[][] roads) {
        int ans = 0;
        for (int mask = 0; mask < 1 << n; ++mask) {
            int[][] g = new int[n][n];
            for (var e : g) {
                Arrays.fill(e, 1 << 29);
            }
            for (var e : roads) {
                int u = e[0], v = e[1], w = e[2];
                if ((mask >> u & 1) == 1 && (mask >> v & 1) == 1) {
                    g[u][v] = Math.min(g[u][v], w);
                    g[v][u] = Math.min(g[v][u], w);
                }
            }
            for (int k = 0; k < n; ++k) {
                if ((mask >> k & 1) == 1) {
                    g[k][k] = 0;
                    for (int i = 0; i < n; ++i) {
                        for (int j = 0; j < n; ++j) {
                            g[i][j] = Math.min(g[i][j], g[i][k] + g[k][j]);
                        }
                    }
                }
            }
            int ok = 1;
            for (int i = 0; i < n && ok == 1; ++i) {
                for (int j = 0; j < n && ok == 1; ++j) {
                    if ((mask >> i & 1) == 1 && (mask >> j & 1) == 1) {
                        if (g[i][j] > maxDistance) {
                            ok = 0;
                        }
                    }
                }
            }
            ans += ok;
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(2^n × (n^3 + m)
Space
O(n^2)

Approach Breakdown

SORT + SCAN
O(n log n) time
O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION
O(n) time
O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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